Problem 29 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

You are constructing a closed rectangular box with a square base and a volume of 200 cubic inches. If the material for the base and lid costs 10 cents per square inch and the material for the sides costs 7 cents per square inch, express the cost of material for the box as a function of \(s\), the length of the side of the base.

Step-by-Step Solution

Answer

The cost of the material for the box as a function of \(s\) is \(C(s) = 0.20s^2 + 56/s\).

Step 1: Find the height of the box

As it is given that the box is a rectangular box with a square base and volume of 200 cubic inches, the volume \((V)\) can be expressed as \(V = s^2 \cdot h\) where \(s\) is the length of the sides of the base and \(h\) is the height of the box. Since the volume of this box is 200 cubic inches, we can rearrange this formula to solve for height, \(h\), as: \(h = V/s^2 = 200/s^2\).

Step 2: Find the surface area

Then the surface area (A) of the box can be determined by the formula \(A = 2(s^2) + 4(s \cdot h)\). Substitute \(h = 200/s^2\) into the surface area formula, the equation will resolve as \(A = 2s^2 + 4s(200/s^2) = 2s^2 + 800/s\).

Step 3: Find the cost of materials

Next, calculate the cost of the materials for the box. The material for the base and lid costs 10 cents per square inch and is \(2s^2\). So the cost for the base and lid rounded to the nearest cent, would be \(0.10 \cdot 2s^2\). While the cost for the sides is 7 cents per square inch and is \(800/s\) square inches, being rounded to the nearest cent would be \(0.07 \cdot 800/s\).

Step 4: Formulate the function

Finally, the total cost (C) is the sum of the cost of the materials for the base, sides, and lid. Therefore, the cost of materials for the box as a function of \(s\) can be expressed as \(C(s) = 0.10 \cdot 2s^2 + 0.07 \cdot 800/s = 0.20s^2 + 56/s\).

Key Concepts

Volume of a Rectangular Box

When we talk about the volume of a rectangular box, we refer to the amount of space inside it. This space is often measured in cubic units, such as cubic inches, cubic feet, or cubic meters. For a box with a square base, like the one in our exercise, the volume is found by multiplying the area of the base by the height of the box. The formula for calculating the volume of a box with a square base is given by \(V = s^2 \cdot h\), where \(s\) represents the side length of the base, and \(h\) stands for the height of the box.

In the textbook exercise, we needed to find the height, given a volume of 200 cubic inches. Knowing the volume formula, we can easily isolate and calculate the height as \(h = \frac{V}{s^2} = \frac{200}{s^2}\). This step is crucial because it serves as a foundation for further calculations in determining the surface area and the overall cost of materials.

Surface Area Calculation

The surface area of an object is the total area that the surface of the object occupies. For our rectangular box, the surface area includes the area of the base, top (which also has the same area as the base since it's a square), and the four sides. The generalized formula for the surface area of a rectangular box is \(A = 2l\cdot w + 2l\cdot h + 2w\cdot h\), where \(l\), \(w\), and \(h\) are the length, width, and height of the box respectively.

In the case of a box with a square base, \(l\) and \(w\) are equal to \(s\), simplifying the formula to \(A = 2s^2 + 4s \cdot h\). After substituting the height we previously calculated, the formula becomes \(A = 2s^2 + 800/s\), which we will ultimately use to compute the cost of the material required for making the box.

Cost Function in Calculus

A cost function in calculus is a mathematical model that represents the cost of producing goods as a function of variable inputs. In the context of our problem, we use a cost function to express the total cost of materials for creating a box in terms of the side length of the square base (\(s\)).

After calculating the surface area required for the base, top, and sides, we need to determine the cost of those materials per square inch. The base and lid cost 10 cents per square inch, leading to a cost of \(0.10 \cdot 2s^2\) for both pieces. Similarly, the sides cost 7 cents per square inch, amounting to \(0.07 \cdot 800/s\) for all four sides. Thus, the total cost function, combining both components, comes out as \(C(s) = 0.20s^2 + 56/s\), which gives us a direct relationship between the size of the base and the total cost.

Quadratic Equations

A quadratic equation is a polynomial equation of the second degree, typically in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents a variable. These equations are fundamental in many areas of mathematics and can be solved by various methods such as factoring, completing the square, or using the quadratic formula.

In our exercise, the cost function \(C(s) = 0.20s^2 + 56/s\) includes terms that resemble a quadratic equation when looking at the \(s^2\) term. However, because of the \(56/s\) term, it's not a standard quadratic equation. Even so, if we were to set this cost function equal to a certain cost, we could rearrange the terms to a standard quadratic form and then apply methods such as factoring or the quadratic formula to solve for \(s\), showing the interconnectedness of these mathematical concepts.